Meaning of an element "vanishing" on $V(I)$ I am reading a question on Vakil's algebraic geometry notes and I am confused by a particular term (exercise 3.4.J in the November 17, 2018 draft on page 118).
Suppose $I\subset B$ is an ideal where $B$ is a commutative ring. We are asked to "Show that $f$ vanishes on $V(I)$ if and only if $f\in\sqrt{I}$".
My question is what does it mean for $f\in B$ to
vanish" on $V(I)$? I know we sometimes say an element of a ring vanishes at a prime ideal if it is contained in the prime ideal. However, what does it mean in this context?
 A: This means $f$ belongs to all prime ideals in $\operatorname{Spec}B$ that contain $I$. As the intersection of all these prime ideals is the root of the ideal $I$…
A: I'm not familiar with Vaki'l book, but usually $f=f(x_1,\ldots,x_n)$ is a polynomial over a field $k$ and $V(I)$ is a subset of $\mathbb A^n=k^n.$ Then all this means is $f(a_1,\ldots,a_n)=0$ for all $(a_1,\ldots, a_n)\in V(I).$

Edit by tkf:
Although not answering the question, this does provide intuition for the correct answer.  We may regard elements of $\mathbb A^n$ as primes in the ring $B=k[x_1,\ldots,x_n]$.  Specifically $a=(a_1,\ldots,a_n)$ corresponds to the prime $p_a=\langle x_1-a_1,\ldots,x_n-a_n\rangle$.  Then in this case given $f\in B$, we have equivalent statements: $$f\in p_a \iff f(a_1,\ldots,a_n)=0.$$
If we let $V_I\subseteq \mathbb A^n$ consist of elements $a$ such that $g(a)=0$ for all $g\in I$, then for all $a\in \mathbb A^n$:$$p_a\supseteq I  \iff a\in V_I.$$
For a general commutative ring $B$, the notions on the right which this user used to define $f$ vanishing on $V_I$ do not make sense, but the notions on the left do, leading to the correct explanation given by @Bernard.
